Chemistry Solutions

When more solute can be dissolved in a solution at a constant temperature, it is an unsaturated solution.

When no more solute can be dissolved in a solution at a constant temperature and pressure, it is called a saturated solution.

Kindly consider the following figure

Mixture of acetone and chloroform shows negative deviation from Raoult's law because of hydrogen bonding

  $P_T=P_A^o{X}_A+P_B^o{X}_B$

$170=200X_A+100\left(1-X_A\right)$

$170=200X_A+100-100X_A$  

X_A = 0.7                   X_B = 0.3

$Y_A=\frac{P_A^o{X}_A}{P_T}=\frac{0.7*200}{170}=0.82$ f

Y_A = p_a /p_a + p_b = 400 x ½ / 400 x ½ + 500 x ½ = 4/9

Higher the value of (ixM), higher will be the boiling point of solution.

Applying conservation of momentum for collision of blocks

$2mv+0=3m{v}^{\text{'}}$

$?v^{\text{'}}=\frac{2v}{3}\dots \left(1\right)$

Now, from conservation of energy

$\frac{1}{2}*3m*{v^{\text{'}}}^2=\frac{1}{2}k{x}_0^2$

$x_0=\sqrt[]{\frac{3m{v}^{\text{'}2}}{k}}=v\text{'}\sqrt[]{\frac{3m}{k}}$

$=\frac{2v}{3}\sqrt[]{\frac{3m}{k}}=v\sqrt[]{\frac{4m}{3k}}$

$A_2{B}_3?2A^{+3}+3B^{-2}$

123

$\therefore i=1+4\alpha =1+4*0.6=1+2.4=3.4$

$\Delta T_b=i{k}_{b}m=3.4*0.52*1=1.768\approx 1.77K$

$T_b=374.77K\approx 375K.$

$\Delta T_b=i{K}_{b}m$

$\Delta T_f=i{K}_{f}m$

$\therefore \frac{4}{4}=\frac{K_b*1.5}{K_f*4.5}$

$\frac{K_b}{K_f}=3$